柯西不等式正实数x+y+z=1,求证x的4次幂/﹙2+y²-z﹚+y的4次幂/﹙2+z²-x﹚+z的4次幂/﹙2+x²-y﹚≥1/48在线等重谢
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![柯西不等式正实数x+y+z=1,求证x的4次幂/﹙2+y²-z﹚+y的4次幂/﹙2+z²-x﹚+z的4次幂/﹙2+x²-y﹚≥1/48在线等重谢](/uploads/image/z/1775553-33-3.jpg?t=%E6%9F%AF%E8%A5%BF%E4%B8%8D%E7%AD%89%E5%BC%8F%E6%AD%A3%E5%AE%9E%E6%95%B0x%2By%2Bz%3D1%2C%E6%B1%82%E8%AF%81x%E7%9A%844%E6%AC%A1%E5%B9%82%2F%EF%B9%992%2By%26%23178%3B-z%EF%B9%9A%2By%E7%9A%844%E6%AC%A1%E5%B9%82%2F%EF%B9%992%2Bz%26%23178%3B-x%EF%B9%9A%2Bz%E7%9A%844%E6%AC%A1%E5%B9%82%2F%EF%B9%992%2Bx%26%23178%3B-y%EF%B9%9A%E2%89%A51%2F48%E5%9C%A8%E7%BA%BF%E7%AD%89%E9%87%8D%E8%B0%A2)
柯西不等式正实数x+y+z=1,求证x的4次幂/﹙2+y²-z﹚+y的4次幂/﹙2+z²-x﹚+z的4次幂/﹙2+x²-y﹚≥1/48在线等重谢
柯西不等式正实数x+y+z=1,
求证x的4次幂/﹙2+y²-z﹚+y的4次幂/﹙2+z²-x﹚+z的4次幂/﹙2+x²-y﹚≥1/48在线等重谢
柯西不等式正实数x+y+z=1,求证x的4次幂/﹙2+y²-z﹚+y的4次幂/﹙2+z²-x﹚+z的4次幂/﹙2+x²-y﹚≥1/48在线等重谢
利用柯西不等式
[X^4/(2+y^2-z) +y^4/(2+z^2-x) +z^4/(2+x^2-y)] (2+y^2-z +2+z^2-x +2+x^2-y)>=(x^2+y^2+z^2) 显然x=y=z时等号成立
由于x+y+z=1,带入第二个括号(2+y^2-z +2+z^2-x +2+x^2-y),得
[X^4/(2+y^2-z) +y^4/(2+z^2-x) +z^4/(2+x^2-y)] (5+x^2+y^2+z^2)>=(x^2+y^2+z^2)
整理上述不等式可得:
[X^4/(2+y^2-z)+y^4/(2+z^2-x)+z^4/(2+x^2-y)]
>=(x^2+y^2+z^2)/(5+x^2+y^2+z^2)=1--5/(5+x^2+y^2+z^2)
即[X^4/(2+y^2-z)+y^4/(2+z^2-x)+z^4/(2+x^2-y)]>=1--5/(5+x^2+y^2+z^2)
再由柯西不等式可得(x^2+y^2+z^2)(1+1+1)>=(x+y+z)^2暨(x^2+y^2+z^2)>=1/3
等号成立当且仅当x=y=z
再把(x^2+y^2+z^2)>=1/3带入此式:[X^4/(2+y^2-z)+y^4/(2+z^2-x)+z^4/(2+x^2-y)]>=1--5/(5+x^2+y^2+z^2)可得该式>=1/48
两个大于等于号的成立条件都是x=y=z,故而这个连续不等式是可以的.
所以[X^4/(2+y^2-z)+y^4/(2+z^2-x)+z^4/(2+x^2-y)]>=1/48